X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y =

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

          =

          =

    Let the slope of the tangent which is perpendicular to this line is ''

    Then by the property of perpendicular lines,

   Now the equation of the line passing though (2, 10) having slope

           y - y' =

           y - 10 =

           y - 10 =

                  y =

                  y =

Therefore, equation of the line will be, y =